How do you find the domain and range of f(x)= sqrt(x^4-16x^2)?

Sep 25, 2015

Domain $\left\{x \in \mathbb{R} , x = 0 , x \ge 4 \mathmr{and} x \le - 4\right\}$

Range $\left\{y \in \mathbb{R} , y \ge 0\right\}$

Explanation:

Write $f \left(x\right) = \sqrt{{x}^{2} \left(x - 4\right) \left(x + 4\right)}$

For f(x) to be real , either $x = 0 , \mathmr{and} \left(x - 4\right) \left(x + 4\right) \ge 0$

This implies that either both x-4 and x+4 should be$\ge 0 \mathmr{and} \le 0$
which means either $x \ge 4 \mathmr{and} x \le - 4$.

Hence domain would be $\left\{x \in \mathbb{R} , x = 0 , x \ge 4 \mathmr{and} x \le - 4\right\}$. In iterval notation it would be $\left(- \infty , - 4\right] U 0 U \left[4 , \infty\right)$

For range it is clear that for x=0, y=0 and for $x \ge 4 \mathmr{and} x \le - 4$, y would be positive. Thus range would be $\left\{y \in \mathbb{R} , y \ge 0\right\}$